Explanation

(a)
Let $I = \int_{}^{}\frac{e^{x}}{\left( 2 + e^{x} \right)(e^{x} + 1)}dx$
Put $e^{x} = t\ \Rightarrow \ e^{x}dx = dt$
$\therefore\ I = \int_{}^{}\frac{dt}{(2 + t)(t + 1)}$
$= \int_{}^{}\left\lbrack \frac{1}{(1 + t)} - \frac{1}{(2 + t)} \right\rbrack dt$
$= \log{(1 + t)} - \log{(2 + t)} + c$
$= \log\left( \frac{1 + t}{2 + t} \right) + c$
$= \log\left( \frac{1 + e^{x}}{2 + e^{x}} \right) + c$